Question 59156
Does it equals zero? In that case...


{{{ 105a^3b+27a^2b^2-33ab^3 = 3ab (35a^2+9ab-11b^2) = 0 }}}


From here (from the first multiplier) we can see that one solution is {{{a = 0}}}, another is {{{b = 0}}}.


What solutions are there from the second multiplier (from the second brackets)?
If {{{ b <> 0 }}}, we can divide both sides by {{{ b^2 }}}:


{{{ 35(a/b)^2 + 9(a/b) - 11 = 0 }}}


Here we get a quadratic equation, let the quadratic equation solver solve this for us:


*[invoke solve_quadratic_equation 35, 9, -11]


So, {{{ a/b = (-9 + sqrt (1621)) / 70}}} or {{{ a/b = (-9 - sqrt (1621)) / 70}}}
or
{{{ a = b(-9 + sqrt (1621)) / 70}}} or {{{ a = b(-9 - sqrt (1621)) / 70}}}


Meaning that you can take very (infinitly) many different "a"s and "b"s as long as they have one of these two proportions.

So, along with the two zero cases (at the beginning), this is the solution of the given equation.